Question

If $\mathit{\sigma }\mathbf{\in }{\mathit{S}}_{\mathbf{n}}$is a product of disjoint transpositions, prove that${\mathit{\sigma }}^{\mathbf{2}}\mathbf{=}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$ .

Short answer

It is proved that,${\sigma }^2=\left(1\right)$ .

Step-by-step solution

Determine σ2=(1)

Consider given that $\sigma \in S_n$is a product of disjoint transpositions.

Assume that role="math" localid="1654605184640" $\sigma ={\tau }_1{\tau }_2\mathrm{.....}{\tau }_k$for some disjoint transpositions ${\tau }_1,{\tau }_2,\mathrm{....},{\tau }_k\in S_n$.

If $\alpha ,\beta \in S_n$are two disjoint permutations then, $\alpha \beta =\beta \alpha$.

Consequently, for $1\le i\ne j\le k$, ${\tau }_i$and ${\tau }_j$are disjoint.

Therefore, ${\tau }_i{\tau }_j={\tau }_j{\tau }_i$.

Find ${\sigma }^2$as:

$$\begin{gathered} {\sigma }^2={\left({\tau }_1\mathrm{....}{\tau }_k\right)}^2 \\ ={\tau }_1\mathrm{....}{\tau }_k{\tau }_1\mathrm{....}{\tau }_k \\ ={\tau }_1{\tau }_1\mathrm{.....}{\tau }_k{\tau }_k \\ ={{\tau }_1}^2\mathrm{....}{{\tau }_k}^2 \end{gathered}$$

This implies,${\sigma }^2=e$ .

Here, the last equality holds as for any transpositions$\tau$ .

Therefore,${\tau }^2=e$ .

Hence, it is proved that ${\sigma }^2=\left(1\right)$.